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Question

In triangle $$ABC$$; $$\angle ABC=90^{o},\angle CAB=x^{o},\tan x^{o}=\dfrac {3}{4}$$ and $$BC=15\ cm$$. Find the measure of $$AB$$ and $$AC$$.


Solution

R.E.F.Image.
$$Tanx = 3/4$$
$$ BC/AB = 3/4$$
$$ 15/AB=3/4$$
$$ 3AB = 60$$
$$ AB = 20 cm $$
In right $$\Delta ABC $$
$$ AC^{2} = AB^{2}+Bc^{2} $$  [by Pythagoras]
$$ = 20^{2}+15^{2}$$
$$ = 400+225 = 625$$
$$ AC = \sqrt{625} = 25 cm$$
$$ \therefore AB = 20 cm ; AC = 25 cm.$$

1207038_1283627_ans_bc9507090b0e4901b5e73f09e1d2aab2.jpg

Maths

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